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Question
If A \[\begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\] , then show that |3 A| = 27 |A|.
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Solution
\[A = \begin{bmatrix}1 & 0 & 1 \\ 0 & 1 & 2 \\ 0 & 0 & 4\end{bmatrix}\]
\[ \Rightarrow 3A = \begin{bmatrix}3 & 0 & 3 \\ 0 & 3 & 6 \\ 0 & 0 & 12\end{bmatrix} \left[\text{ Multiplying each element of A by 3 }\right]\]
\[ \Rightarrow \begin{vmatrix}3A\end{vmatrix} = \left( - 1 \right)^{1 + 1} 3\left( 36 - 0 \right) + \left( - 1 \right)^{1 + 2} 0\left( 0 - 0 \right) + \left( - 1 \right)^{1 + 3} 3\left( 0 - 0 \right) = 3\left( 36 - 0 \right) - 0\left( 0 - 0 \right) + 3\left( 0 - 0 \right) \left[\text{ Expanding along }R_1 \right]\]
\[ = 3 \times 36 = 108 . . . \left( 1 \right)\]
\[ \Rightarrow \begin{vmatrix}A\end{vmatrix} = \left( - 1 \right)^{1 + 1} 1\left( 4 - 0 \right) + \left( - 1 \right)^{1 + 2} 0\left( 0 - 0 \right) + \left( - 1 \right)^{1 + 3} 1\left( 0 - 0 \right) = 1\left( 4 - 0 \right) - 0\left( 0 - 0 \right) + 1\left( 0 - 0 \right) = 4 \left[\text{ Expanding along }R_1 \right]\]
\[ \Rightarrow 27\left| A \right| = 27 \times 4 = 108 . . . \left( 2 \right)\]
\[ \therefore \left| 3A \right| = 27 \left| A \right| \left[\text{ From eqs . (1) and (2) }\right]\]
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